A 110. G of sample of copper (copper = 0.385 J/g·°C) is heated to 82.4°C and then placed in an insulated container of water (water = 4.18 J/g·°C) at 22.3°C. The final temperature of the water and copper is 24.9°C. What is the volume of water? (water = 1 g/mL)

Volume of water is 224mL

Explanation:

The heat that sample of copper reducing its temperature is the same heat that is absorbed from the water using the equation:

C(Cu)*m*ΔT = -C(water)*m*ΔT

Where C is specific heat of Copper and water, m is the mass and ΔT is change in temperature.

We need to solve the mass of water and as density of water is 1g/mL the mass in grams is equal to volume in mL

The copper change from 82.4°C to 24.9°C = 57.5°C

Mass of copper = 110g

Specific heat of copper and water is 0.385J/g°C and 4.18J/g°C, respectively.

Mass of water is our incognite and change in temperature of water is 22.3°C - 24.9°C = -2.6°C

Replacing:

C(Cu)*m*ΔT = -C(water)*m*ΔT

0.385J/g°C*110g*57.5°C = -4.18J/g°C*m*-2.6°C

224g = Mass of water

That means the volume of water is 224mL